Help: Math Homework and I’m stuck! Explanation below. (Solve 2y- 3(2y- 3) = 2y - 5(3y - 4))

[yoyleberriex]

Directions are to solve each equation (solve for x or whatever variable it may be.) The particular problem I was having trouble on is
2y- 3(2y- 3) = 2y - 5(3y - 4)
I simplify it down to y = 1.33 (repeating) but once I check work my answer is incorrect and I can’t figure out where I went wrong. Please help!

 

[stapel]

Directions are to solve each equation (solve for x or whatever variable it may be.) The particular problem I was having trouble on is
2y- 3(2y- 3) = 2y - 5(3y - 4)
I simplify it down to y = 1.33 (repeating) but once I check work my answer is incorrect and I can’t figure out where I went wrong. Please help!

Did you try checking if your solution is correct? The left-hand side evaluates as:

[imath]\qquad 2\left(\frac{4}{3}\right) - 3\left(2\left(\frac{4}{3}\right) - 3\right)[/imath]

[imath]\qquad \frac{8}{3} - 3\left(\frac{8}{3} - \frac{9}{3}\right)[/imath]

[imath]\qquad \frac{8}{3} - 3\left(-\frac{1}{3}\right)[/imath]

[imath]\qquad \frac{8}{3} + \frac{3}{3}[/imath]

[imath]\qquad \frac{11}{3}[/imath]

The right-hand side evaluates as:

[imath]\qquad 2\left(\frac{4}{3}\right) - 5\left(3\left(\frac{4}{3}\right) - 4\right)[/imath]

[imath]\qquad \frac{8}{3} - 5(4 - 4)[/imath]

[imath]\qquad \frac{8}{3} - 5(0)[/imath]

[imath]\qquad \frac{8}{3}[/imath]

Since the two sides do not evaluate to the same number, then [imath]y = \frac{4}{3}[/imath] cannot be a solution to the equation.

Unfortunately, since we cannot see your work, there is no way for us to help you find your error. Kindly please reply with that information.

 

[TristanH]

Directions are to solve each equation (solve for x or whatever variable it may be.) The particular problem I was having trouble on is
2y- 3(2y- 3) = 2y - 5(3y - 4)
I simplify it down to y = 1.33 (repeating) but once I check work my answer is incorrect and I can’t figure out where I went wrong. Please help!

Let’s start with distribution on both sides
2y-(6y+9)=2y-(15y-20)
Simplify
-4y - 9 = -13y + 20
Add 13y
9y - 9 = 20
Add 9
9y = 29
Divide by 9
y = 29/9
29/9 is not equal to 4/3

 

[Dr.Peterson]

Let’s start with distribution on both sides
2y-(6y+9)=2y-(15y-20)
Simplify
-4y - 9 = -13y + 20
Add 13y
9y - 9 = 20
Add 9
9y = 29
Divide by 9
y = 29/9
29/9 is not equal to 4/3

Not quite; you made a sign error. If you check your answer, you find that

[imath]2\cdot\frac{29}{9}-3\left(2\cdot\frac{29}{9}-3\right)=\frac{58}{9}-3\left(\frac{58}{9}-\frac{27}{9}\right)=\frac{58}{9}-3\left(\frac{31}{9}\right)=\frac{58}{9}-\frac{93}{9}=\frac{-65}{9}[/imath]​

[imath]2\cdot\frac{29}{9}-5\left(3\cdot\frac{29}{9}-4\right)=\frac{58}{9}-5\left(\frac{87}{9}-\frac{36}{9}\right)=\frac{58}{9}-5\left(\frac{51}{9}\right)=\frac{58}{9}-\frac{255}{9}=\frac{-197}{9}[/imath]​

So it doesn't work. It's good to add a solution to an abandoned thread, but we have to check our work.

Here is my work:

2y- 3(2y- 3) = 2y - 5(3y - 4)

[imath]2y-3(2y-3) = 2y-5(3y-4)\\2y-(6y-9) = 2y-(15y-20)\\2y-6y+9 = 2y-15y+20\\-4y+9 = -13y+20\\-4y+13y+9 = 20\\9y+9 = 20\\9y = 11\\y = \frac{11}{9}[/imath]​

When I check this, I get

[imath]2\cdot\frac{11}{9}-3\left(2\cdot\frac{11}{9}-3\right)=\frac{22}{9}-3\left(\frac{22}{9}-\frac{27}{9}\right)=\frac{22}{9}-3\left(\frac{-5}{9}\right)=\frac{22}{9}+\frac{15}{9}=\frac{37}{9}[/imath]​

[imath]2\cdot\frac{11}{9}-5\left(3\cdot\frac{11}{9}-4\right)=\frac{22}{9}-5\left(\frac{33}{9}-\frac{36}{9}\right)=\frac{22}{9}-5\left(\frac{-3}{9}\right)=\frac{22}{9}+\frac{15}{9}=\frac{37}{9}[/imath]​

So this solution works.

 

[lookagain]

Let’s start with distribution on both sides
2y-(6y+9)=2y-(15y-20)
...

TristanH (and Dr.Peterson),

As 2y - 3(2y - 3) = 2y - 5(3y - 4) is equivalent to 2y + [-3(2y - 3)] = 2y + [-5(3y - 4)],
the point of distributing should also involve the clearing of pertinent grouping
symbols at the same time, as you get to distribute -3 and -5, respectively.
Here is a redo:

\(\displaystyle 2y - 3(2y - 3) \ = \ 2y - 5(3y - 4)\)

\(\displaystyle 2y - 6y + 9 \ = \ 2y - 15y + 20\)

\(\displaystyle -4y + 9 \ = \ -13y + 20\)

\(\displaystyle 13y - 4y \ = \ 20 - 9\)

\(\displaystyle 9y \ = \ 11 \)

\(\displaystyle y \ = \ \dfrac{11}{9} \)


Again, this candidate solution would be checked in the original equation.

 

[Dr.Peterson]

TristanH (and Dr.Peterson),

As 2y - 3(2y - 3) = 2y - 5(3y - 4) is equivalent to 2y + [-3(2y - 3)] = 2y + [-5(3y - 4)],
the point of distributing should also involve the clearing of pertinent grouping
symbols at the same time, as you get to distribute -3 and -5, respectively.
Here is a redo:

\(\displaystyle 2y - 3(2y - 3) \ = \ 2y - 5(3y - 4)\)

\(\displaystyle 2y - 6y + 9 \ = \ 2y - 15y + 20\)

\(\displaystyle -4y + 9 \ = \ -13y + 20\)

\(\displaystyle 13y - 4y \ = \ 20 - 9\)

\(\displaystyle 9y \ = \ 11 \)

\(\displaystyle y \ = \ \dfrac{11}{9} \)


Again, this candidate solution would be checked in the original equation.

There are, of course, more than one way to handle the details, which in part depend on what a student is ready for.

For myself, I would probably have subtracted 2y from both sides first, and perhaps multiplied by -1 before doing anything else. But since I tell beginners to simplify each side before solving, I wouldn't do that in showing a struggling student the process.

If I hadn't done that, I myself would distribute the -3 and -5 in their entirety as you did, but since some students aren't ready to see a subtraction as adding a negative, I chose to take it in two steps.

I just wish we could see what mistake the OP made, so we could focus on that.

 

[TristanH]

Not quite; you made a sign error. If you check your answer, you find that

[imath]2\cdot\frac{29}{9}-3\left(2\cdot\frac{29}{9}-3\right)=\frac{58}{9}-3\left(\frac{58}{9}-\frac{27}{9}\right)=\frac{58}{9}-3\left(\frac{31}{9}\right)=\frac{58}{9}-\frac{93}{9}=\frac{-65}{9}[/imath]​

[imath]2\cdot\frac{29}{9}-5\left(3\cdot\frac{29}{9}-4\right)=\frac{58}{9}-5\left(\frac{87}{9}-\frac{36}{9}\right)=\frac{58}{9}-5\left(\frac{51}{9}\right)=\frac{58}{9}-\frac{255}{9}=\frac{-197}{9}[/imath]​

So it doesn't work. It's good to add a solution to an abandoned thread, but we have to check our work.

Here is my work:

[imath]2y-3(2y-3) = 2y-5(3y-4)\\2y-(6y-9) = 2y-(15y-20)\\2y-6y+9 = 2y-15y+20\\-4y+9 = -13y+20\\-4y+13y+9 = 20\\9y+9 = 20\\9y = 11\\y = \frac{11}{9}[/imath]​

When I check this, I get

[imath]2\cdot\frac{11}{9}-3\left(2\cdot\frac{11}{9}-3\right)=\frac{22}{9}-3\left(\frac{22}{9}-\frac{27}{9}\right)=\frac{22}{9}-3\left(\frac{-5}{9}\right)=\frac{22}{9}+\frac{15}{9}=\frac{37}{9}[/imath]​

[imath]2\cdot\frac{11}{9}-5\left(3\cdot\frac{11}{9}-4\right)=\frac{22}{9}-5\left(\frac{33}{9}-\frac{36}{9}\right)=\frac{22}{9}-5\left(\frac{-3}{9}\right)=\frac{22}{9}+\frac{15}{9}=\frac{37}{9}[/imath]​

So this solution works.

Whoops! I must’ve accidentally included the minus with the 3 on the outside with the negative 3 on the inside!